Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

In the oxidation of ethane: 2 C2H6 + 7 02 → 4 CO2 + 6 H2O how many
liters of O2 are required to react with 90 grams of ethane?

Sagot :

Eduard22sly

Answer:

10.5 L of O₂.

Explanation:

We'll begin by calculating the number of mole in 90 g of C₂H₆. This can be obtained as follow:

Molar mass of C₂H₆ = (12×2) + (6×1)

= 24 + 6

= 30 g/mol

Mass of C₂H₆ = 90 g

Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 90 / 30

Mole of C₂H₆ = 3 moles.

Finally, we shall determine the volume of O₂ required. This is illustrated:

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above, we can say that:

2 L of C₂H₆ required 7 L of O₂.

Therefore, 3 L of C₂H₆ will require

= (3 × 7)/ 2 = 10.5 L of O₂.

Thus, 10.5 L of O₂ is required for the reaction.

We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.